Theorem B.2.27Let f : K → ℝqbe continuous at every point of K where Kis a closed and bounded set in ℝp. Then f is uniformly continuous.

Proof: Suppose not. Then there exists ε > 0 and sequences

{xj}

and

{yj}

of points in K
such that

|xj − yj| < 1
j

but

|f (xj) − f (yj)|

≥ ε. Then Theorem 4.7.2 which says K is sequentially compact, there is a
subsequence

{xnk}

of

{xj}

which converges to a point x ∈ K. Then since

|xnk − ynk|

<

1
k

, it
follows that

{ynk}

also converges to x. Therefore,

ε ≤ kli→m∞|f (xnk)− f (ynk)| = |f (x)− f (x)| = 0,

a contradiction. Therefore, f is uniformly continuous as claimed. ■

Later in the book, I will consider the fundamental theorem of algebra. However, here
is a fairly short proof based on the extreme value theorem. You may have to fill
in a few details however. In particular, note that

(ℂ,|⋅|)

is the same as

(ℝ2,|⋅|)

where

|⋅|

is the standard norm on ℝ2. Thus closed and bounded sets are compact in

(ℂ, |⋅|)

.

Proposition B.2.28Let p

(z)

= a0 + a1z +

⋅⋅⋅

+ an−1zn−1 + znbe a nonconstantpolynomial where each ai∈ ℂ. Then there is a root to this polynomial.

Proof: Suppose the nonconstant polynomial p

(z)

= a0 + a1z +

⋅⋅⋅

+ zn, has no zero in
ℂ. Since lim

|z|

→∞

|p (z)|

= ∞, there is a z0 with

|p(z0)| = min |p(z)| > 0
z∈ℂ

Why? (The growth condition shows that you can restrict attention to a closed and bounded
set and then apply the extreme value theorem.) Then let q

(z)

=

p(zp+(zz00))

. This is also a
polynomial which has no zeros and the minimum of

|q(z)|

is 1 and occurs at z = 0. Since
q

(0)

= 1, it follows q

(z)

= 1 + akzk + r

(z)

where r

(z)

consists of higher order terms. Here ak
is the first coefficient which is nonzero. Choose a sequence, zn→ 0, such that akznk< 0. For
example, let −akznk =